0
$\begingroup$

I was working with a rank-2 tensor $G^{\mu \nu}$ and exploring the effect of covariant differentiation on it, until I found that it satisfied a "covariant eigenvalue equation" of the form: $$\partial_\mu G^{\mu \nu} = k_\mu G^{\mu \nu}$$ where $k_\mu$ is not a constant but some function of the coordinates $x^\mu$. I found an ansatz that could work for it: $G^{\mu \nu}=B^{\mu \nu} \exp{\Big( \int k_\alpha dx^\alpha \Big)}$ with $B^{\mu \nu}$ a constant tensor. In my specific case, $k_\mu$ is explicitly described as $-\partial_\mu (\ln\sqrt{-g})$ where $g=\det{g_{\mu \nu}}$, but was doubtful that the "four-antiderivative" of that expression simplified to something like $-\ln\sqrt{-g}$. Is there a more elegant way of going about my differential equation in general?

$\endgroup$
  • $\begingroup$ Seems to me like the ansatz you have given does not depend on x. Rather, "x" appears as a dummy variable in the integration \int{k.dx} in the exponential. In this case not only is B a constant, but so is G. Therefor, dG=0. So, it is a solution, but trivial, and only for k=0. $\endgroup$ – hft Jun 18 '18 at 1:00
  • $\begingroup$ I was sloppy with my notation; I should have specified that my "four-antiderivative" is actually something like $\int_{x^\alpha _0} ^{x^\alpha } k_\alpha ds^\alpha$, such that for two variables it would look like $\int_{x_0} ^x k_x dx' + \int_{y_0} ^y k_y dy'$ $\endgroup$ – N.E. Jun 18 '18 at 1:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.